TITLE: Prove that the symmetric group $S_n$, $n \geq 3$, has trivial center.
QUESTION [5 upvotes]: I am trying to prove this:

Let $\sigma$ be a non-identity element of $S_{n}$. If $n \geq 3$ show that $\exists \gamma \in  S_{n}$ such that $\sigma\gamma \neq \gamma\sigma$.
Hint: Let $\sigma*k=1$ where $k \neq 1$. Let $m$ not be an element of $\{k,1\}$. Let $\gamma=(k,m)$

Here's the answer from the textbook:
Let $\sigma*k=1$ for some $k \neq 1$. Then as $m\geq 3$, choose an $m$ not an element of $\{k,1\}$. Now let $\gamma=(k,m)$. This gives $\gamma*\sigma*k=\gamma*(1)=1$ but $\sigma*\gamma*k=\sigma*m \neq 1$, since if $\sigma*m=1=\sigma*k$, then $m=k$, as $\sigma$ is one-to-one contrary to the assumption.
I guess I'm just not understanding the steps to their proof. Can anybody make sense of this to someone new to proofs? I mostly don't understand how $\sigma*\gamma*k=\sigma*m=1$

REPLY [2 votes]: Consider two permutations
$\sigma =(a$ $ b)$
$\gamma =(b$ $a$ $c$ $\cdots \cdots$)
Note gamma can be of any length $(n\geq 3)$.
$\sigma \circ \gamma (a) = c$
$\gamma \circ \sigma (a) = a$
Thus $\sigma \circ \gamma \neq \gamma \circ \sigma$
Thus no $two-cycle$ commutes with all the elements. Also, no $n-cycle$ commutes with all the elements.